Question: What is the extraneous solution to these equations? $\dfrac{x^2 - 2x}{x - 8} = \dfrac{x + 40}{x - 8}$
Explanation: Multiply both sides by $x - 8$ $ \dfrac{x^2 - 2x}{x - 8} (x - 8) = \dfrac{x + 40}{x - 8} (x - 8)$ $ x^2 - 2x = x + 40$ Subtract $x + 40$ from both sides: $ x^2 - 2x - (x + 40) = x + 40 - (x + 40)$ $ x^2 - 2x - x - 40 = 0$ $ x^2 - 3x - 40 = 0$ Factor the expression: $ (x - 8)(x + 5) = 0$ Therefore $x = 8$ or $x = -5$ At $x = 8$ , the denominator of the original expression is 0. Since the expression is undefined at $x = 8$, it is an extraneous solution.